MOIST AIR (Psychrometry)

“ MANAGEMENT HANDBOOK” sixth edition, Chapter 6, 10 and 17

Psychrometry is the science of studying the thermodynamic properties of moist air and the use of these properties to analyze conditions and processes involving moist air, for example • • processes • gas

Composition of Air

Ratio compared to Dry Air (%) Molecular - M Boiling Point Gas Chemical Symbol (kg/kmol) (oC) By By weight

Oxygen 20.95 23.20 32.00 O2 -196

Nitrogen 78.09 75.47 28.02 N2 -183

Carbon Dioxide 0.03 0.046 44.01 CO2 -78.5

Hydrogen 0.00005 ~ 0 2.02 H2 -252.87

Argon 0.933 1.28 39.94 Ar -186

Neon 0.0018 0.0012 20.18 Ne -246

Helium 0.0005 0.00007 4.00 He -269

Krypton 0.0001 0.0003 83.8 Kr -153.4

Xenon 9 10-6 0.00004 131.29 Xe -108.1 The Law • p V = m R T (1) •where • p = absolute (Pa) • V= volume of gas (m3) • m = mass (kg) R = individual gas constant (J/kg,K) T = absolute (K) • This equation (1) can be modified to: • p = ρ R T (2) • where the • ρ = m / V (3) The individual gas constant can be expressed with the universal gas constant and the molecular weight of the air as: R = Ru / Mgas (2) where Mgas = molecular weight of the gas Ru = 8314.51 = universal gas constant (J/(kmol.K)) Dry air is more dense than moist air.

Individual Gas Constant Molecular - R Gas Weight SI Units (kg/kmol) (J/kg.K) Dry Air 286.9 28.97 455 18.02 vapor

Pressure in Moist Air - Daltons Law

Daltons Law for moist air can be expressed as: p = pa + pw (1) where p = total pressure of air (Pa) pa = dry air (Pa) pw = partial pressure (Pa) The maximum pressure possible before vapor start to condensate at an actual temperature is called the saturation pressure - pws. can be expressed as: pws = e(77.3450 + 0.0057 T - 7235 / T) / T^8.2 (1) where pws = water vapor saturation pressure (Pa) e = the constant 2.718...... T = temperature of the moist air (K) or using table (page 7 in document)

o Temperature ( C) Water Vapor Saturation Pressure (Pa)

0 609.9

5 870

10 1225

15 1701

20 2333

25 3160

30 4234

Example - The Saturation Pressure of Water Vapor The Saturation pressure of water vapor in moist air at 25oC can be calculated as: pws = e( 77.3450 + 0.0057 (273.15 + 25) - 7235 / (273.15 + 25) ) / (273.15 + 25)^8.2= 3160 Pa Relative (φ)

Relative humidity is the ratio of the water - pw , to the vapor pressure of saturated air at the same temperature - pws, expressed as a percentage.

• Relative humidity by partial pressure: • φ = pw / pws 100% where • φ = relative humidity (%) • pw = vapor pressure (Pa) • pws=saturation vapor pressure at the actual dry bulb temperature (Pa) Example - Relative Humidity and Vapor Pressure

From table the saturation pressure at 20°C is 2337 Pa. If the vapor pressure in the actual air is 1636 Pa, the relative humidity can be calculated as: 1636 φ =⋅=100 70% RH 2337 Humidity Ratio (x) • Humidity ratio is the actual mass of water vapor present in moist air - to the mass of the dry air. Humidity ratio is normally expressed in kilogram water vapor per kilogram dry air (SI-units). • The specific humidity or humidity ratio can be expressed as: • x = mw / ma (1) where • x = specific humidity or humidity ratio (kg water/kg air) • mw = mass of water vapor (kg) • ma = mass of the dry air (kg) The humidity ratio can also be expressed with the partial pressure of water vapor and air: • x = 0.622 pw / (pt - pw) •where • pw = partial pressure of water vapor in the moist air (Pa) • pa = partial pressure of the moist air (Pa) total pressure • (pt - pw) is the partial pressure of dry air • The maximum amount of water vapor in the air is achieved when pw = pws the saturation pressure of water vapor at the actual temperature. (2) can be modified to: • xs = 0.622 pws / (pt - pws) •where • xs = humidity ratio at saturation (kgwater/kgair) • pws = saturation pressure of water vapor

Note! The water vapor capacity increases dramatically with temperature. This is important especially in drying processes. Example - Specific Humidity of Moist Air

The specific humidity for saturated humid air at 20oC with water vapor partial pressure 2337 Pa at of 101.325 kPa can be calculated as: • x = 0.622*2337 / (101325 - 2337 ) • = 0.0147 (kg/kg) •= 14.7g/kg Example – Density of Moist Air

What density have the air in the earlier example.

1 kg of dry air “have” also 0.0147 kg water vapor. For the water vapor the ideal gas gives R mT⋅ u ⋅ w M p ==2337 w w V 8314.5 0.0147⋅⋅ 293.15 Vm==18.02 0.853 3 2337 The moist air density mm+ 1.000+ 0.0147 ρ ===dry air water vapor 1.190kg / m3 V 0.853 (h)

Enthalpy - h - (kJ/kg) of moist air is defined as the total enthalpy of the dry air and the water vapor mixture per kg of moist air, includes the

• enthalpy of the dry air - the sensible - and • the enthalpy of the evaporated water - the • Enthalpy of moist air can be expressed as: • h = ha + x hw where • h = specific enthalpy of moist air (kJ/kg) • ha = specific enthalpy of dry air (kJ/kg) • x = humidity ratio (kg/kg) • hw = specific enthalpy of water vapor (kJ/kg)

h = Cpaww⋅ t + x [h +Cp⋅ t] kJ/kg

h = 1.01⋅ t + x [2502+1.84⋅ t] kJ/kg Example - Enthalpy in Moist Air

The enthalpy of humid air at 25oC with specific moisture content x = 0.0203 kg/kg, can be calculated as: • h = 1.01 25 + 0.0203 [2502 + 1.84 25] • = 25.25 + 51.72 •= 76.97(kJ/kg of dry air) • Note! The latent heat due to of water is the major part of the enthalpy. Moist air

p ϕ= w ⋅ 100(% RH) pws

pp xk = 0.622ww =0.622 (/)gkg ppaw−−101325 pw h = 1.01⋅⋅ t + x [2502+1.84 t] (kJ/kg)

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